picture of solution for when you multiply first and add after
link of original discussion of the picture: Combined geometric and arithmetic series partial sum
In this picture, you can see a way of solving combined arithmetic and geometric sequence, however for this equation to work you need to multiply first and then add after. I want to be able to do it the other way around.
I know hoe to this somewhat by doing the following:
=((+)⋅+)⋅
You can just keep adding: { ⋅+) } to the equation to find the next term but this is still somewhat tedious, I want to find a way in which you can just find the nth term by plugging in its positional value into the equation. Meaning if I want to find the 5th term n=5 and so on.
- the nth term, for example, the 5th term of the sequence
- The original value, or the first value in the series
- the amount added every time
- the amount multiplied / common ration I think
If anybody knows how to do this that would be amazing Thank you!
$$\begin{align} a_n &= \bigg( \Bigl( \bigl( (a_1 + d) \cdot q + d \bigr) \cdot q + d \Bigr) \cdot q + d \bigg) \cdot q + \cdots \\ \\ &= \Bigl( \bigl( (a_1q + dq + d) \cdot q + d \bigr) \cdot q + d \Bigr) \cdot q + \cdots \\ \\ &= \bigl( (a_1q^2 + dq^2 + dq + d) \cdot q + d \bigr) \cdot q + \cdots \\ \\ &= (a_1q^3 + dq^3 + dq^2 + dq + d) \cdot q + \cdots \\ \\ & \;\; \vdots \\ \\ &= a_1q^{n-1} + dq^{n-1} + dq^{n-2} + \cdots + dq \\ \\ &= a_1q^{n-1} + dq \; (q^{n-2} + q^{n-3} + \cdots + 1) \\ \\ &= \boxed{ a_1q^{n-1} + dq \left( \frac{q^{n-1}-1}{q-1} \right) } \end{align}$$